\(1,=\sqrt{6+4\sqrt{2}}-\sqrt{11-6\sqrt{2}}=2+\sqrt{2}-\left(3-\sqrt{2}\right)=2\sqrt{2}-1\\ 2,=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}=2\sqrt{5}-1+2\sqrt{5}+1=4\sqrt{5}\\ 3,=\sqrt{\dfrac{14-6\sqrt{5}}{2}}=\dfrac{3-\sqrt{5}}{\sqrt{2}}=\dfrac{3\sqrt{2}-\sqrt{10}}{2}\\ 4,=0\\ 5,=\sqrt{2}\sqrt{2-\sqrt{3}}\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\\ =\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\\ =8+4\sqrt{3}-4\sqrt{3}-6=2\\ 6,=\sqrt{15-4\sqrt{14}}-\sqrt{15-4\sqrt{14}}=0\\ 7,=\dfrac{-\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}=-\sqrt{2}\\ 8,=\dfrac{\sqrt{2}-\sqrt{3}}{-\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)}=-\dfrac{1}{\sqrt{2}}=\dfrac{-\sqrt{2}}{2}\)
